2022 H2 Maths Paper 1 Q6 — Worked Solution

Part (a)

4 marks

Approach

$y = \frac{1}{x}$ is transformed by translations and a stretch. To see these directly, we first rewrite $f(x) = \frac{ax + k}{x - a}$ in the form $\frac{c}{x - a} + a$ by polynomial division, which matches the shape of a transformed $\frac{1}{x}$ .

Writing $ax + k = a(x - a) + \left( k + a^{2} \right)$ , since $a(x - a) = ax - a^{2}$ leaves remainder $k + a^{2}$ ,

f(x) = \frac{a(x - a) + \left( k + a^{2} \right)}{x - a}

= a + \frac{k + a^{2}}{x - a}

To check your answer: expanding

a + \frac{k + a^{2}}{x - a} = \frac{a(x - a) + k + a^{2}}{x - a} = \frac{ax - a^{2} + k + a^{2}}{x - a} = \frac{ax + k}{x - a}

, matching the original

f(x)

✓

Starting from $y = \frac{1}{x}$ , the curve $y = f(x) = \frac{k + a^{2}}{x - a} + a$ is obtained by:

Step 1: a translation of $a$ units in the positive $x$ -direction (replacing $x$ with $x - a$ ), giving $y = \frac{1}{x - a}$ .

Step 2: a stretch parallel to the $y$ -axis, scale factor $\left( a^{2} + k \right)$ , giving $y = \frac{a^{2} + k}{x - a}$ .

Step 3: a translation of $a$ units in the positive $y$ -direction, giving $y = \frac{a^{2} + k}{x - a} + a = f(x)$ .

$\therefore$ $y = \frac{1}{x}$ is translated $a$ units in the positive $x$ -direction, stretched parallel to the $y$ -axis with scale factor $a^{2} + k$ , then translated $a$ units in the positive $y$ -direction.

Part (b)

2 marks

Let $y = \frac{ax + k}{x - a}$ . Solving for $x$ in terms of $y$ ,

y(x - a) = ax + k

\Rightarrow xy - ay = ax + k

\Rightarrow xy - ax = k + ay

\Rightarrow x(y - a) = k + ay

\Rightarrow x = \frac{ay + k}{y - a}

Swapping $x$ and $y$ ,

f^{- 1}(x) = \frac{ax + k}{x - a}

To check your answer:

f^{- 1}(x)

has exactly the same formula as

f(x)

. Testing with

a = 2

,

k = 3

,

x = 5

:

f(5) = \frac{13}{3}

, and

f\left( \frac{13}{3} \right) = 5

, so

f

maps

5 \rightarrow \frac{13}{3} \rightarrow 5

, confirming

f

is self-inverse ✓

$\therefore$ $f^{- 1}(x) = \frac{ax + k}{x - a} = f(x)$ .

Part (c)

1 mark

f^{2}(x) = x

Explanation: by (b), $f^{- 1}(x) = f(x)$ , so $f$ is an involution (self-inverse). Applying $f$ twice returns the original input, $f^{2}(x) = f\left( f(x) \right) = f\left( f^{- 1}(x) \right) = x$ .

Part (d)

2 marks

Approach

since $f^{2}(x) = x$ for every $x$ , applying $f$ an even number of times is the identity, and applying it an odd number of times is the same as applying it once. As $2023$ is odd, $f^{2023}$ reduces to a single application of $f$ .

Since $2022$ is even, $f^{2022}(y) = y$ for any $y$ , so

f^{2023}(x) = f^{2022}\left( f(x) \right)

= f(x)

Hence $f^{2023}(1) = f(1)$ .

f(1) = \frac{a(1) + k}{1 - a}

= \frac{a + k}{1 - a}

To check your answer: testing with

a = 2

,

k = 3

: direct iteration of

f

starting from

1

, applied

2023

times, gives

- 5

, matching

\frac{a + k}{1 - a} = \frac{2 + 3}{1 - 2} = \frac{5}{- 1} = - 5

✓

$\therefore$ $f^{2023}(1) = \frac{a + k}{1 - a}$ .

2022 H2 Maths Paper 1, Question 6

Part (a)

Part (b)

Part (c)

Part (d)

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